I'm self studying some course material I found online and got stuck in an exercise about a binary relation $R$ on a set $X$ with $|X|=6$. The reader is asked to prove that there exist different $x,y,z\in X$ for which either $(x,y),(y,z),(z,x)\in R$ or $(x,y),(y,z),(z,x)\notin R.$ Also an example for showing that the property does not hold without symmetry is asked.
EDIT: I misunderstood the question. A friend explained what is actually asked here. The task is to show that there exists a cycle of three elements of the set $X$ (if the elements are thought of as nodes of an undirected graph and the relation as edges between them) in the relation $R$ or outside of it. I'm having a hard time trying to prove it, any help is still very welcomed.
There are $2\times{6\choose2}=30$ possible two-element pairs for the symmetric relation $R$ and ${6\choose3}=20$ combinations of three different elements from the set $X$. Is there always a cycle of three elements in one of the partitions of the set into pairs in $R$ and pairs not in $R$?
EDIT 2: Progress on the first part: turned out that the amount of potential edges is not sufficient condition for the property to hold. $6$ edges ($12$ pairs for the relation $R$) can be chosen without a cycle and choosing them leaves $9$ edges ($18$ potential pairs for the relation $R$) outside. A cycle requires $3$ nodes and $3$ edges. Both $6$ and $9$ are divisible by $3$. On the other hand, if $|X|=5$, the numbers for edges in and potential deges outside are both $5$, which is not divisible by $3$ and also the property does not hold with $|X|=5$. With $|X|<5$ there are not enough potential edges outside the relation so the property cannot hold. I'll continue from the lead on divisibility.